Math 104: Introduction to Analysis Summer 2019
Lecture 9: Subsequences, lim sup and lim inf
Lecturer: Michael Christianson 9 July Aditya Sengupta
9.1
More about subsequences
For the midterm, it is important to understand a proof of the following: Proposition 9.1. If lim
n→+∞sn exists, then every subsequence of (sn) has a limitn→+∞lim sn.
To prove this, we showed that for any (snk), nk ≥ k∀k. Then we applied the definition of a limit.
Proposition 9.2. If (sn) is not bounded above, there exists a subsequence (snk) diverging to +∞.
Proof. ∀k, pick nk such that snk does what we want, i.e. cause snk to be big when k is big. The simplest
choice here is snk ≥ k. Then k is not an upper bound on {sn | n ∈ N}, that is, there exists nk such that
snk > k. Now apply the definition of a limit to show that lim
k→+∞snk = +∞.
Important statements that may be used in other proofs, but which we won’t have to prove: Theorem 9.3. (Bolzano-Weierstrass) Every bounded sequence has a convergent subsequence. Theorem 9.4. Let (sn) be a sequence, and let S be the set of all subsequential limits.
(a) lim sup sn, lim inf sn ∈ S.
(b) sup S = lim sup sn, inf S = lim inf sn
(c) lim
n→+∞sn exists iff S contains one element.
Proof. (c) If lim
n→+∞sn exists, then every subsequence has limitn→+∞lim sn, therefore S = { limn→+∞sn}. If S
contains one element, lim sup sn= lim inf sn, therefore lim
n→+∞sn exists
Example 9.1. Let sn= 1 + (−1)n, tn= n cos nπ3 .
(a) What are the subsequences of (sn) and (tn)?
(s2k)k∈N = (2, 2, 2, . . . ), which converges to 2, and (s2k+1)k∈N = (0, 0, 0, . . . ),
so S = {0, 2} is the set of all subsequential limits.
For all subsequences (snk) of (sn), either infinitely many of nk are even or in-
finitely many of nk are odd. So, there exists a subsequence of (snk) converging
to either 0 or 2. If lim
Lecture 9: Subsequences, lim sup and lim inf 47
lim
n→+∞snk∈ {0, 2}.
For the subsequences of tn, we do not see any constant terms; any subsequence
grows or decays to ±∞. Since tn is not bounded above or below, we see that
±∞ are subsequential limits. So T = {, ±∞}. (b) What are lim sup and lim inf of both?
lim sup sn = sup S = 2 and lim inf sn = inf S = 0; lim sup tn = sup T = +∞
and lim inf tn= inf T = −∞.
(c) Prove (sn) and (tn) have no limit.
The sets S and T have more than one element, but this is not the case if lim
n→+∞sn orn→+∞lim tn exist.
(d) Which of (sn) and (tn) are bounded?
We expect that (sn) is bounded and (tn) is not. If a sequence is not bounded
above, we expect that there exists a subsequence such that the limit of that subsequence is +∞, i.e. +∞ is in the set of subsequential limits. The same holds for being bounded below and the limit being −∞. Therefore, because ±∞ ∈ T , (tn) is not bounded, and because ±∞ /∈ S, (sn) is not bounded.
9.2
Using sequence properties
We can show that the rational numbers are countable in the form of a sequence argument, Proposition 9.5. There exists a sequence (rn)n∈N such that ∀q ∈ Q, ∃!n s. t. rn= q
Proof. Cantor diagonalization. We can extend this to the following,
Proposition 9.6. ∀a ∈ R, > 0, ∃ infinitely many q ∈ Q s. t. q ∈ (a − , a + ). Therefore the set
S= {q ∈ Q | |q − a| < } ≡ {n ∈ N | |rn− a| < }
is infinite (there is a bijection between the two sets). Therefore, there exists a subsequence of (rn) converging
to a. For all k, there exists an rnk ∈ Q such that a −
1
n < rnk < a by denseness of Q. So every real number is
9.3
lim sup and lim inf
Recall that for a sequence (sn), if it is bounded above, its lim sup is lim sup sn = lim
N →∞(sup{sn | n > N })
and lim inf is defined similarly if it is bounded below. If (sn) is not bounded above, lim sup sn= +∞ and if
it is not bounded below, lim inf sn= −∞.
Let lN = inf{sn| n > N } and uN = sup{sn | n > N }. We see that lN ≤ uN∀N , and therefore lim N →+∞lN ≤
lim
N →+∞uN, so lim inf sn ≤ lim sup sn.
For all N , {sn | n > N + 1} ⊆ {sn | n > N }, therefore uN +1 ≤ uN and similarly lN +1 ≥ lN. So uN is
decreasing, i.e. lim
N →+∞exists, and lN is increasing, soN →+∞lim lN exists.
lim sup sn= lim
N →+∞uN ≤ u1= sup{sn | n > 1} < +∞
Therefore if (sn) is bounded above, lim sup sn∈ R or is −∞. Similarly, if (sn) is bounded below, lim inf sn≥
l1= inf{sn| n > 1} > −∞.
Theorem 9.7. For all decreasing sequences (sn), lim
n→+∞sn= inf{sn | n ∈ N}. For all increasing sequences
(sn), lim
n→+∞sn= sup{sn| n ∈ N}.
An alternative way of dealing with lim sup is
lim sup sn = lim
N →+∞uN = inf{uN} = inf{sup{sn| n > N } | N ∈ N}
lim inf sn = lim
N →+∞lN = sup{inf{sn| n > N } | N ∈ N}
Theorem 9.8. Let (sn) be a sequence. Then, lim
n→+∞sn exists iff lim sup sn = lim inf sn and in that case
lim sup sn= lim inf sn= lim n→+∞sn
Intuitively, it is important to understand that lim sup sn and lim inf sn are like lim
n→+∞sn but are always
defined.
Theorem 9.9. Let (sn) be a sequence and let S be the set of its subsequential limits. Then
(a) lim sup sn, lim inf sn ∈ S
(b) sup S = lim sup sn, inf S = lim inf sn
(c) If for all subsequences snk of (sn) such that its limit exists, we have lim
k→+∞snk ≤ α ∈ R, then
lim sup sn≤ α.
This theorem has been written in lecture like three times, so I may go back and clean this up.